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<div class=”textbox learning-objectives”>
<h3>Learning Objectives</h3>
<ul>
<li>Determine whether a given ordered pair is a solution of a given linear equation.</li>
<li>Find solutions of a linear equation.</li>
<li>Complete a table of solutions.</li>
</ul>
</div>
<div class=”textbox key-takeaways”>
<h3>Key words</h3>
<ul>
<li><strong>Ordered pair solution</strong>: a solution written in the form [latex]\left (x,y\right )[/latex]</li>
</ul>
</div>
<h2>Finding Solutions of Linear Equations in Two Variables</h2>
When an equation has two variables, any solution will be an ordered pair with a value for each variable.
<div id=”post-378″ class=”standard post-378 chapter type-chapter status-publish hentry”>
<div class=”entry-content”>
<div class=”textbox shaded”>
<h3>Solution to a Linear Equation in Two Variables</h3>
An ordered pair [latex]\left(x,y\right)[/latex] is a solution of the linear equation [latex]ax+by=c[/latex], if the equation is a true statement when the [latex]x[/latex]– and [latex]y[/latex]-values of the ordered pair are substituted into the equation.
</div>
<div class=”bcc-box bcc-info”>
<h3>Example</h3>
Determine whether [latex](−2,4)[/latex] is a solution of the equation [latex]4y+5x=3[/latex].
<h4>Solution</h4>
Substitute [latex]x=−2[/latex] and [latex]y=4[/latex] into the equation:
<p style=”text-align: center;”>[latex]\begin{array}{r}4y+5x=3\\4\left(4\right)+5\left(−2\right)=3\end{array}[/latex]</p>
Evaluate.
<p style=”text-align: center;”>[latex]\begin{array}{r}16+\left(−10\right)=3\\6=3\end{array}[/latex]</p>
The statement is not true, so [latex](−2,4)[/latex] is not a solution.
<h4>Answer</h4>
[latex](−2,4)[/latex] is not a solution of the equation [latex]4y+5x=3[/latex].
</div>
</div>
</div>
<div class=”textbox exercises”>
<h3>example</h3>
Determine which ordered pairs are solutions of the equation [latex]x+4y=8\text{:}[/latex]
1. [latex]\left(0,2\right)[/latex]
2. [latex]\left(2,-4\right)[/latex]
3. [latex]\left(-4,3\right)[/latex]
<h4>Solution</h4>
Substitute the [latex]x\text{- and }y\text{-values}[/latex] from each ordered pair into the equation and determine if the result is a true statement.
<table id=”eip-id1168469838906″ class=”unnumbered unstyled” summary=”This image shows three columns. The first column is labeled “>
<tbody>
<tr>
<td>1. [latex]\left(0,2\right)[/latex]</td>
<td>2. [latex]\left(2,-4\right)[/latex]</td>
<td>3. [latex]\left(-4,3\right)[/latex]</td>
</tr>
<tr>
<td>[latex]x=\color{blue}{0}, y=\color{red}{2}[/latex][latex]x+4y=8[/latex]
[latex]\color{blue}{0}+4\cdot\color{red}{2}\stackrel{?}{=}8[/latex]
[latex]0+8\stackrel{?}{=}8[/latex]
[latex]8=8\checkmark[/latex]</td>
<td>[latex]x=\color{blue}{2}, y=\color{red}{-4}[/latex][latex]x+4y=8[/latex]
[latex]\color{blue}{2}+4(\color{red}{-4})\stackrel{?}{=}8[/latex]
[latex]2+(-16)\stackrel{?}{=}8[/latex]
[latex]-14\not=8[/latex]</td>
<td>[latex]x=\color{blue}{-4}, y=\color{red}{3}[/latex][latex]x+4y=8[/latex]
[latex]\color{blue}{-4}+4\cdot\color{red}{3}\stackrel{?}{=}8[/latex]
[latex]-4+12\stackrel{?}{=}8[/latex]
[latex]8=8\checkmark[/latex]</td>
</tr>
<tr>
<td>[latex]\left(0,2\right)[/latex] is a solution.</td>
<td>[latex]\left(2,-4\right)[/latex] is not a solution.</td>
<td>[latex]\left(-4,3\right)[/latex] is a solution.</td>
</tr>
</tbody>
</table>
</div>
<div class=”textbox key-takeaways”>
<h3>try it</h3>
<iframe id=”ohm146928″ class=”resizable” src=”https://ohm.lumenlearning.com/multiembedq.php?id=146928&theme=oea&iframe_resize_id=ohm146928&show_question_numbers” width=”100%” height=”150″></iframe>
</div>
<div class=”textbox exercises”>
<h3>example</h3>
Determine which ordered pairs are solutions of the equation. [latex]y=5x - 1\text{:}[/latex]
1. [latex]\left(0,-1\right)[/latex]
2. [latex]\left(1,4\right)[/latex]
3. [latex]\left(-2,-7\right)[/latex]
<h4>Solution</h4>
Substitute the [latex]x\text{-}[/latex] and [latex]y\text{-values}[/latex] from each ordered pair into the equation and determine if it results in a true statement.
<table id=”eip-id1168466112848″ class=”unnumbered unstyled” summary=”The figure shows three algebraic substitutions in three columns, a, b, and c. The equation is y equals 5 times x minus 1. The first substitution is for ordered pair (0, -1). The first line in column a reads x = 0, with 0 shown in blue and y = -1, with -1 shown in red. The next line is y = 5 x – 1. The next line is -1, shown in red = 5 open parentheses 0, shown in blue, closed parentheses – 1, with a question mark shown over the equal sign. The next line is -1 = 0 – 1, with a question mark over the equal sign. The next line is – 1 = -1, followed by a check mark. The last line is “>
<tbody>
<tr>
<td>1. [latex]\left(0,-1\right)[/latex]</td>
<td>2. [latex]\left(1,4\right)[/latex]</td>
<td>3. [latex]\left(-2,-7\right)[/latex]</td>
<td>[latex]x=\color{blue}{0}, y=\color{red}{-1}[/latex][latex]y=5x-1[/latex]
[latex]\color{red}{-1}\stackrel{?}{=}5(\color{blue}{0})-1[/latex]
[latex]-1\stackrel{?}{=}0-1[/latex]
[latex]-1=-1\checkmark[/latex]</td>
<td>[latex]x=\color{blue}{1}, y=\color{red}{4}[/latex][latex]y=5x-1[/latex]
[latex]\color{red}{4}\stackrel{?}{=}5(\color{blue}{1})-1[/latex]
[latex]4\stackrel{?}{=}5-1[/latex]
[latex]4=4\checkmark[/latex]</td>
<td>[latex]x=\color{blue}{-2}, y=\color{red}{-7}[/latex][latex]y=5x-1[/latex]
[latex]\color{red}{-7}\stackrel{?}{=}5(\color{blue}{-2})-1[/latex]
[latex]-7\stackrel{?}{=}-10-1[/latex]
[latex]-7\not=-11[/latex]</td>
</tr>
<tr>
<td>[latex]\left(0,-1\right)[/latex] is a solution.</td>
<td>[latex]\left(1,4\right)[/latex] is a solution.</td>
<td>[latex]\left(-2,-7\right)[/latex] is not a solution.</td>
</tr>
</tbody>
</table>
</div>
</div>
</div>
<div class=”textbox key-takeaways”>
<h3>try it</h3>
<iframe id=”ohm146929″ class=”resizable” src=”https://ohm.lumenlearning.com/multiembedq.php?id=146929&theme=oea&iframe_resize_id=ohm146929&show_question_numbers” width=”100%” height=”150″></iframe>
<iframe id=”ohm146941″ class=”resizable” src=”https://ohm.lumenlearning.com/multiembedq.php?id=146941&theme=oea&iframe_resize_id=ohm146941&show_question_numbers” width=”100%” height=”150″></iframe>
</div>
The video shows more examples of how to determine whether an ordered pair is a solution of a linear equation.
<iframe src=”https://www.youtube.com/embed/9aWGxt7OnB8?feature=oembed&rel=0″ width=”500″ height=”281″ frameborder=”0″ allowfullscreen=”allowfullscreen”></iframe>
<h2>Complete a Table of Solutions</h2>
In the previous examples, we substituted the [latex]x\text{- and }y\text{-values}[/latex] of a given ordered pair to determine whether or not it was a solution of a given linear equation. But how do we find the ordered pairs if they are not given? One way is to choose a value for [latex]x[/latex] and then solve the equation for [latex]y[/latex]. Or, choose a value for [latex]y[/latex] and then solve for [latex]x[/latex].
Let’s consider the equation [latex]y=5x - 1[/latex]. The easiest value to choose for [latex]x[/latex] or [latex]y[/latex] is zero:
[latex]\begin{equation}\begin{aligned}y & =5x-1 \;\;\;\;\;\;\;\;\;\;\text{Substitute}\;x=0\\y & = 5(0)-1\\y & = -1\end{aligned}\end{equation}[/latex] So, [latex]x=0,\;y=-1[/latex] is a solution, which as an ordered pair is [latex]\left (0,\,-1\right )[/latex].
[latex]\begin{equation}\begin{aligned}y & =5x-1 \;\;\;\;\;\;\;\;\;\text{Substitute}\;y=0\\0 & = 5x-1\;\;\;\;\;\;\;\;\;\text{Solve for}\;x\\1 & = 5x\\ \frac{1}{5} & =x\end{aligned}\end{equation}[/latex] So, [latex]x=\frac{1}{5},\;y=0[/latex] is a solution, which as an ordered pair is [latex]\left (\frac{1}{5},\,0\right )[/latex].
We can continue to find more solutions by choosing different values of [latex]x[/latex] and [latex]y[/latex].
Suppose [latex]x=2[/latex]:
<table id=”eip-id1168469817063″ class=”unnumbered” summary=”The figure shows a substitution into an equation and accompanying comments. The first equation is y = 5 open parentheses 2, shown in blue, closed parentheses – 1. The comment is “>
<tbody>
<tr>
<td></td>
<td>[latex]y=5x - 1[/latex]</td>
</tr>
<tr>
<td>Substitute [latex]x=2[/latex]</td>
<td>[latex]y=5(\color{blue}{2})-1[/latex]</td>
</tr>
<tr>
<td>Multiply.</td>
<td>[latex]y=10 - 1[/latex]</td>
</tr>
<tr>
<td>Simplify.</td>
<td>[latex]y=9[/latex]</td>
</tr>
</tbody>
</table>
To find a third solution, we’ll let [latex]x=2[/latex] and solve for [latex]y[/latex].
We can write our solutions in a table:
<table id=”fs-id1569134″ class=”unnumbered” summary=”This table has 5 rows and 3 columns. The first row is the equation y = 5 x – 1. The next row is a header row and it labels each column “>
<thead>
<tr valign=”top”>
<th colspan=”3″>[latex]y=5x - 1[/latex]</th>
</tr>
<tr valign=”top”>
<th>[latex]x[/latex]</th>
<th>[latex]y[/latex]</th>
<th>[latex]\left(x,y\right)[/latex]</th>
</tr>
</thead>
<tbody>
<tr valign=”top”>
<td>[latex]0[/latex]</td>
<td>[latex]-1[/latex]</td>
<td>[latex]\left(0,-1\right)[/latex]</td>
</tr>
<tr valign=”top”>
<td>[latex]\frac{1}{5}[/latex]</td>
<td>[latex]0[/latex]</td>
<td>[latex]\left(\frac{1}{5},0\right)[/latex]</td>
</tr>
<tr valign=”top”>
<td>[latex]2[/latex]</td>
<td>[latex]9[/latex]</td>
<td>[latex]\left(2,9\right)[/latex]</td>
</tr>
</tbody>
</table>
We can find more solutions to the equation by substituting any value of [latex]x[/latex] or any value of [latex]y[/latex] and solving the resulting equation to get another ordered pair that is a solution. There are an infinite number of solutions for this equation.
<div class=”textbox exercises”>
<h3>example</h3>
Complete the table to find three solutions of the equation [latex]y=4x - 2\text{:}[/latex]
<table id=”fs-id1599948″ class=”unnumbered” summary=”This table has 5 rows and 3 columns. The first row is the equation y = 4 x – 2. The next row is a header row and it labels each column “>
<thead>
<tr valign=”top”>
<th colspan=”3″>[latex]y=4x - 2[/latex]</th>
</tr>
<tr valign=”top”>
<th>[latex]x[/latex]</th>
<th>[latex]y[/latex]</th>
<th>[latex]\left(x,y\right)[/latex]</th>
</tr>
</thead>
<tbody>
<tr valign=”top”>
<td>[latex]0[/latex]</td>
<td></td>
<td></td>
</tr>
<tr valign=”top”>
<td>[latex]-1[/latex]</td>
<td></td>
<td></td>
</tr>
<tr valign=”top”>
<td>[latex]2[/latex]</td>
<td></td>
<td></td>
</tr>
</tbody>
</table>
<h4>Solution</h4>
Substitute [latex]x=0,x=-1[/latex], and [latex]x=2[/latex] into [latex]y=4x - 2[/latex].
<table id=”eip-id1168468326216″ class=”unnumbered unstyled” summary=”The figure shows three algebraic substitutions into an equation. The first substitution x = 0, with 0 shown in blue. The next line is y = 4 x – 2. The next line is y = 4 times 0, shown in blue, minus 2. The next line is y = 0 – 2. The next line is y = -2. The last line is “>
<tbody>
<tr>
<td>[latex]x=\color{blue}{0}[/latex]</td>
<td>[latex]x=\color{blue}{-1}[/latex]</td>
<td>[latex]x=\color{blue}{2}[/latex]</td>
</tr>
<tr>
<td>[latex]y=4x - 2[/latex]</td>
<td>[latex]y=4x - 2[/latex]</td>
<td>[latex]y=4x - 2[/latex]</td>
</tr>
<tr>
<td>[latex]y=4\cdot{\color{blue}{0}}-2[/latex]</td>
<td>[latex]y=4(\color{blue}{-1})-2[/latex]</td>
<td>[latex]y=4\cdot{\color{blue}{2}}-2[/latex]</td>
</tr>
<tr>
<td>[latex]y=0 - 2[/latex]</td>
<td>[latex]y=-4 - 2[/latex]</td>
<td>[latex]y=8 - 2[/latex]</td>
</tr>
<tr>
<td>[latex]y=-2[/latex]</td>
<td>[latex]y=-6[/latex]</td>
<td>[latex]y=6[/latex]</td>
</tr>
<tr>
<td>[latex]\left(0,-2\right)[/latex]</td>
<td>[latex]\left(-1,-6\right)[/latex]</td>
<td>[latex]\left(2,6\right)[/latex]</td>
</tr>
</tbody>
</table>
The results are summarized in the table.
<table id=”fs-id1572080″ class=”unnumbered” summary=”This table has 5 rows and three columns. The first row is the equation y = 4 x – 2. The next row is a header row and it labels each column “>
<thead>
<tr valign=”top”>
<th colspan=”3″>[latex]y=4x - 2[/latex]</th>
</tr>
<tr valign=”top”>
<th>[latex]x[/latex]</th>
<th>[latex]y[/latex]</th>
<th>[latex]\left(x,y\right)[/latex]</th>
</tr>
</thead>
<tbody>
<tr valign=”top”>
<td>[latex]0[/latex]</td>
<td>[latex]-2[/latex]</td>
<td>[latex]\left(0,-2\right)[/latex]</td>
</tr>
<tr valign=”top”>
<td>[latex]-1[/latex]</td>
<td>[latex]-6[/latex]</td>
<td>[latex]\left(-1,-6\right)[/latex]</td>
</tr>
<tr valign=”top”>
<td>[latex]2[/latex]</td>
<td>[latex]6[/latex]</td>
<td>[latex]\left(2,6\right)[/latex]</td>
</tr>
</tbody>
</table>
</div>
<div class=”textbox key-takeaways”>
<h3>try it</h3>
<iframe id=”ohm146945″ class=”resizable” src=”https://ohm.lumenlearning.com/multiembedq.php?id=146945&theme=oea&iframe_resize_id=ohm146945&show_question_numbers” width=”100%” height=”150″></iframe>
<iframe id=”ohm146947″ class=”resizable” src=”https://ohm.lumenlearning.com/multiembedq.php?id=146947&theme=oea&iframe_resize_id=ohm146947&show_question_numbers” width=”100%” height=”150″></iframe>
</div>
<div class=”textbox exercises”>
<h3>example</h3>
Complete the table to find three solutions to the equation [latex]5x - 4y=20\text{:}[/latex]
<table id=”fs-id1328205″ class=”unnumbered” style=”width: 479.75px;” summary=”This table is 5 rows and 3 columns. The first row is the equation 5 x – 4 y = 20. The next row is a header row and it labels each column “>
<thead>
<tr valign=”top”>
<th style=”width: 446.75px;” colspan=”3″>[latex]5x - 4y=20[/latex]</th>
</tr>
<tr valign=”top”>
<th style=”width: 114px;”>[latex]x[/latex]</th>
<th style=”width: 114px;”>[latex]y[/latex]</th>
<th style=”width: 218.75px;”>[latex]\left(x,y\right)[/latex]</th>
</tr>
</thead>
<tbody>
<tr valign=”top”>
<td style=”width: 114px;”>[latex]0[/latex]</td>
<td style=”width: 114px;”></td>
<td style=”width: 218.75px;”></td>
</tr>
<tr valign=”top”>
<td style=”width: 114px;”></td>
<td style=”width: 114px;”>[latex]0[/latex]</td>
<td style=”width: 218.75px;”></td>
</tr>
<tr valign=”top”>
<td style=”width: 114px;”></td>
<td style=”width: 114px;”>[latex]5[/latex]</td>
<td style=”width: 218.75px;”></td>
</tr>
</tbody>
</table>
<h4>Solution</h4>
<img src=”https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/277/2017/04/24224816/CNX_BMath_Figure_11_01_038_img.png” alt=”The figure shows three algebraic substitutions into an equation. The first substitution is x = 0, with 0 shown in blue. The next line is 5 x- 4 y = 20. The next line is 5 times 0, shown in blue – 4 y = 20. The next line is 0 – 4 y = 20. The next line is – 4 y = 20. The next line is y = -5. The last line is ” />
The results are summarized in the table.
<table id=”fs-id1572845″ class=”unnumbered” style=”width: 479.75px;” summary=”This table has 5 rows and 3 columns. The first row is equation 5 x – 4 y = 20. The next row is a header row and it labels each column “>
<thead>
<tr valign=”top”>
<th style=”width: 446.75px;” colspan=”3″>[latex]5x - 4y=20[/latex]</th>
</tr>
<tr valign=”top”>
<th style=”width: 114px;”>[latex]x[/latex]</th>
<th style=”width: 114px;”>[latex]y[/latex]</th>
<th style=”width: 218.75px;”>[latex]\left(x,y\right)[/latex]</th>
</tr>
</thead>
<tbody>
<tr valign=”top”>
<td style=”width: 114px;”>[latex]0[/latex]</td>
<td style=”width: 114px;”>[latex]-5[/latex]</td>
<td style=”width: 218.75px;”>[latex]\left(0,-5\right)[/latex]</td>
</tr>
<tr valign=”top”>
<td style=”width: 114px;”>[latex]4[/latex]</td>
<td style=”width: 114px;”>[latex]0[/latex]</td>
<td style=”width: 218.75px;”>[latex]\left(4,0\right)[/latex]</td>
</tr>
<tr valign=”top”>
<td style=”width: 114px;”>[latex]8[/latex]</td>
<td style=”width: 114px;”>[latex]5[/latex]</td>
<td style=”width: 218.75px;”>[latex]\left(8,5\right)[/latex]</td>
</tr>
</tbody>
</table>
</div>
</div>
<div class=”textbox key-takeaways”>
<h3>try it</h3>
<iframe id=”ohm146948″ class=”resizable” src=”https://ohm.lumenlearning.com/multiembedq.php?id=146948&theme=oea&iframe_resize_id=ohm146948&show_question_numbers” width=”100%” height=”150″></iframe>
</div>
<h3></h3>
To find a solution to a linear equation, we can choose any number we want to substitute into the equation for either [latex]x[/latex] or [latex]y[/latex]. We could choose [latex]1,100,-1,000, -\frac{4}{5}, 2.6[/latex], or any other value we want. But it’s a good idea to choose a number that’s easy to work with. We’ll usually choose [latex]0[/latex] as one of our values.
<div class=”textbox exercises”>
<h3>example</h3>
Find a solution to the equation [latex]3x+2y=6[/latex]
<h4>Solution</h4>
<table id=”eip-id1168469770411″ class=”unnumbered unstyled” summary=”The figure shows a four step solution. Step 1 reads “>
<tbody>
<tr>
<td><strong>Step 1:</strong>
Choose any value for one of the variables in the equation.</td>
<td>We can substitute any value we want for [latex]x[/latex] or any value for [latex]y[/latex].Let’s pick [latex]x=0[/latex].
What is the value of [latex]y[/latex] if [latex]x=0[/latex] ?</td>
</tr>
<tr>
<td><strong>Step 2:</strong>
Substitute that value into the equation.Solve for the other variable.</td>
<td>Substitute [latex]0[/latex] for [latex]x[/latex].Simplify.
Divide both sides by [latex]2[/latex].</td>
<td>[latex]3x+2y=6[/latex][latex]3\cdot\color{blue}{0}+2y=6[/latex]
[latex]0+2y=6[/latex]
[latex]2y=6[/latex]
[latex]y=3[/latex]</td>
</tr>
<tr>
<td><strong>Step 3:</strong>
Write the solution as an ordered pair.</td>
<td>So, when [latex]x=0,y=3[/latex].</td>
<td>This solution is represented by the ordered pair [latex]\left(0,3\right)[/latex].</td>
</tr>
<tr>
<td><strong>Step 4:</strong>
Check.</td>
<td>Substitute [latex]x=\color{blue}{0}, y=\color{red}{3}[/latex] into the equation [latex]3x+2y=6[/latex]Is the result a true equation?
Yes!</td>
<td>[latex]3x+2y=6[/latex][latex]3\cdot\color{blue}{0}+2\cdot\color{red}{3}\stackrel{?}{=}6[/latex]
[latex]0+6\stackrel{?}{=}6[/latex]
[latex]6=6\checkmark[/latex]</td>
</tr>
</tbody>
</table>
</div>
<div class=”textbox key-takeaways”>
<h3>try it</h3>
<iframe id=”ohm147000″ class=”resizable” src=”https://ohm.lumenlearning.com/multiembedq.php?id=147000&theme=oea&iframe_resize_id=ohm147000&show_question_numbers” width=”100%” height=”150″></iframe>
</div>
<div class=”textbox key-takeaways”>
<h3>try it</h3>
<iframe id=”ohm147003″ class=”resizable” src=”https://ohm.lumenlearning.com/multiembedq.php?id=147003&theme=oea&iframe_resize_id=ohm147003&show_question_numbers” width=”100%” height=”150″></iframe>
</div>
<div class=”textbox exercises”>
<h3>example</h3>
Find three solutions to the equation [latex]x - 4y=8[/latex].
<h4>Solution</h4>
<table id=”eip-id1168468771120″ class=”unnumbered unstyled” summary=”The figure shows three algebraic substitutions into an equation and accompanying comments. The first starts with the equation x – 4 y = 8. The next line is x = 0, with 0 shown in blue. The next line is 0 – 4 y = 8, with 0 shown in blue. The comment is “>
<tbody>
<tr>
<td>[latex]x-4y=8[/latex]</td>
<td>[latex]x-4y=8[/latex]</td>
<td>[latex]x-4y=8[/latex]</td>
</tr>
<tr>
<td>Choose a value for [latex]x[/latex] or [latex]y[/latex].</td>
<td>[latex]x=\color{blue}{0}[/latex]</td>
<td>[latex]y=\color{red}{0}[/latex]</td>
<td>[latex]y=\color{red}{3}[/latex]</td>
</tr>
<tr>
<td>Substitute it into the equation.</td>
<td>[latex]\color{blue}{0}-4y=8[/latex]</td>
<td>[latex]x-4\cdot\color{red}{0}=8[/latex]</td>
<td>[latex]x-4\cdot\color{red}{3}=8[/latex]</td>
</tr>
<tr>
<td>Solve.</td>
<td>[latex]-4y=8[/latex][latex]y=-2[/latex]</td>
<td>[latex]x-0=8[/latex][latex]x=8[/latex]</td>
<td>[latex]x-12=8[/latex][latex]x=20[/latex]</td>
</tr>
<tr>
<td>Write the ordered pair.</td>
<td>[latex]\left(0,-2\right)[/latex]</td>
<td>[latex]\left(8,0\right)[/latex]</td>
<td>[latex]\left(20,3\right)[/latex]</td>
</tr>
</tbody>
</table>
So [latex]\left(0,-2\right),\left(8,0\right)[/latex], and [latex]\left(20,3\right)[/latex] are three solutions to the equation [latex]x - 4y=8[/latex].
<table id=”fs-id1580614″ class=”unnumbered” summary=”This table it titled x – 4 y =8. It has 4 rows and 3 columns. The first row is a header row and it labels each column “>
<thead>
<tr valign=”top”>
<th colspan=”3″>[latex]x - 4y=8[/latex]</th>
</tr>
<tr valign=”top”>
<th>[latex]x[/latex]</th>
<th>[latex]y[/latex]</th>
<th>[latex]\left(x,y\right)[/latex]</th>
</tr>
</thead>
<tbody>
<tr valign=”top”>
<td>[latex]0[/latex]</td>
<td>[latex]-2[/latex]</td>
<td>[latex]\left(0,-2\right)[/latex]</td>
</tr>
<tr valign=”top”>
<td>[latex]8[/latex]</td>
<td>[latex]0[/latex]</td>
<td>[latex]\left(8,0\right)[/latex]</td>
</tr>
<tr valign=”top”>
<td>[latex]20[/latex]</td>
<td>[latex]3[/latex]</td>
<td>[latex]\left(20,3\right)[/latex]</td>
</tr>
</tbody>
</table>
</div>
Remember, there are an infinite number of solutions to each linear equation. Any ordered pair we find is a solution if it makes the equation true.
<div class=”textbox key-takeaways”>
<h3>TRY IT</h3>
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